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August 31, 2009, 07:57 |
porous inertial resistance
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#1 |
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Hallo,
I want to model an tube bundle heat exchanger. The tube bundles are replaced by a porous media. My problem is, that the porous inertial resistance is a function of to the fluid velocity. Is it possible to create a table, where the porous inertial resistance is shown for x,y,z -components with the according velocity? Star-CCM+ should compute with the porous inertial resistance for the equal velocity from the table. Lukas |
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August 31, 2009, 10:03 |
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#2 |
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TonyD
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Your inertial resistance is in function of your velocity??? OK I have never heard of this but for everything a first time.
When i derive my inertial and viscous factor, i usually have experimental pressure drops vs volume flow rates. Convert these in a plot where you have velocity on the X axis and pressure loss at the Y axis, then use excel or matlab to derive this 2nd degree polyline. Convert according to porous media expression and voila you have your inertial and viscous resistance. dp/dl=Av2+Bv use A and B to find inertial and viscous If now these factors are again dependant of velocity, this would mean a third degree polyline, just derive 3rd degree factors and make sure to use the correct ones to form your inertial and viscous terms. |
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August 31, 2009, 11:44 |
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#3 |
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I calculated the pressure drop with equations from the VDI heat atlas.
The porous inertial resistance can be calculated like this: Pi = dp/dl * 1/vē Is it allowed to remove the porous viscous resistance because of high Re-numbers (Re about 10^5)? dp is a function of velocity, so Pi is a function of velocity, too?! |
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August 31, 2009, 16:45 |
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#4 |
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Aroon
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You can model your inertial (as well as viscous) porous media co-efficient as a function of anything you want (Temp, velocity). Define new Field Function using Tools > Field Function > User Defined Field Function. Name it appropriately and then in Porous Inertial Resistance select Field Function under Method and you can set it to the defined field function.
Hope this helps |
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September 1, 2009, 05:30 |
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#5 |
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Thanks for your answer,
but how I have to declare the field function, that it is according to my table? Lukas |
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September 1, 2009, 10:07 |
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#6 |
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Aroon
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Hello Lukas,
I guess you have a table or a function of Pi = f(V). You can enter this function in the new field function. If you have a table you can always fit a curve and define the new field function based on the defined curve. Regards Aroon |
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September 1, 2009, 10:17 |
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#7 |
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Ok,
it sounds plausible, but Iīm not a expert in using Star-CCM+ Can you give me a step by step manual to do that. I just managed to create a table where the porous inertial resistance is shown for x,y,z -components with the according velocity. But I canīt combine this table with my field function. Lukas |
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September 2, 2009, 10:21 |
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#8 |
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TonyD
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Visharoon,
I dont think he needs resistance in function of velocity, Lukas Pi = dp/dl * 1/vē --> this is your formula, so I have no idea where you get that Pi is in function of v??? Pi is a constant value, the second derative from your 2nd order poly that is dp=f(v) Last edited by bramv101; September 2, 2009 at 11:30. |
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September 2, 2009, 13:15 |
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#9 |
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The problem is ,that my velocity isnīt constant, so my Pi changes, too.
Lukas |
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September 8, 2009, 09:17 |
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#10 |
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TonyD
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Thats the whole idea of a porous media coefficient, your Pi and Pv stay constant while your dp and v change.
for dp/dl=Pi*v^2+Pv*v --> dl, Pi and Pv remain constant Just plot your pressure drop (Y axis) and your velocity (X axis) in excel, you will see that this is a perfect 2nd degree polyline. If you have a Pi dependant on your velocity, that is third degree and this is highly unlikely. Cant imagine such a media.... Inertial = mass*acceleration |
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September 8, 2009, 11:09 |
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#11 |
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This is fact. Iīm aware of this. But the velocity of my fluid, before it permeates the porous media, is between 0 and 2,5 m/s. And for this velocity I have to define my coefficients. If i have a constant velocity before the fluid permeates the porous media, I will have constant coefficients.
Because of the various velocities I want to make the coefficients a function of the velocity. |
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September 8, 2009, 11:35 |
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#12 |
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TonyD
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I understand, but when you say that your velocity goes from 0 to 2.5 [m/s]. Do you mean distributed over the face of the porous media? or is varying in time?
Either way, I still can not find a good enough reason to model your resistance values in function of velocity. In terms of physics this doesn't make sense... Say you can model your dp=f(v) with a third degree poly, what is your goal? |
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September 10, 2009, 07:17 |
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#13 |
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Yes, the velocity is distributed over the face of the porous media. Because of this I think, that I will make a failure, if I calculate one coefficient for this different velocities.
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September 10, 2009, 07:46 |
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#14 |
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TonyD
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Unless your velocities range from 0 to 5000 m/s (unlikely), go for 1 constant value.
If your resistance is constant and your velocity fluctuates then your pressure is a direct result of this. Adapting your resistance to your velocity is like adapting the young modulus of a material to the local force applied -> very wrong So in porous materials inertial and viscous resistance are material properties, these are inherent to the material itself and therefore a material constant. If you have experimental pressure drops vs approach velocity for this type of material just use Darcy to derive Pi and Pv |
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September 11, 2009, 05:24 |
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#15 |
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I want to make the coefficients a function of the velocity outside of the porous media, because this velocity isnīt constant.
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September 11, 2009, 06:15 |
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#16 |
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TonyD
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Could you tell me what your case is a representation off:
* filter applications * geological * solid/gas/liquid * ...... |
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September 11, 2009, 10:54 |
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#17 |
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Ok, I will go for one constant Pi and Pv.
Iīve got a tube bundle heat exchanger with liquid media. The tube bundle should be replaced by a porous media. If I plot the dp/L (y- axis) and v (x-axis) I got a equation like this: y= ax^2+ bx +c a is Pi, b is Pv. Do I have to respect the coefficient c? |
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September 11, 2009, 12:23 |
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#18 |
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TonyD
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you can ignore this c value, usually this is neglible
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September 11, 2009, 12:24 |
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#19 |
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Ok, I have the solution.
Bramv101, thanks for your help. |
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September 11, 2009, 12:24 |
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#20 |
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TonyD
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And to simulate u tube heat exchanger I would go for the heat exchanger tutorial from StarCCM+
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